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Andrew and Katherine rode their bikes and met at a park.
Andrew’s house was 8 miles from the park, Katherine’s house was 5 miles from
the park. They each left their house at the same time and Andrew arrived 10
minutes after Katherine arrived. Katherine’s average speed was 2 mph slower
than Andrew’s average speed. {hint: UNITS matter!}
a. Write an equation to represent this data.
b. What was Andrew’s average speed?

Respuesta :

hadeelalniamia

Answer:

10 min = 1/6 hour

Now we make Andrew's average speed v mph so Katherine's average speed is (v-2) mph

a. 8/v - 5/v-2 = 1/6  [distance = speed x time]

b. 8/v - 5/v-2 = 1/6 , v > 2

   8/v - 5/v-2 = 1/6 = 0

48(v-2) -30v-v-(v-2)/6v(v-2) = 0 [transform the expression]

20v-96-v^2/6v(v-2) = 0

20v-96-v^2 = 0

(v-8)(v-12) = 0

{v1=8}

{v2=12}

Step-by-step explanation:

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